Given point $A$ in the interior of a circle and point $B$ outside the same circle, prove that there is a point in the circle in $\overline{AB}$

Question

I want to run away from continuity/analysis and atempt to prove it using euclidean geometry theorems. I do know that the interior of a circle is a convex region. But I lack a smart way to prove the problem without continuity.

I thought maybe proving that the are other points in the interior of the circle (let's call it $\gamma$) in segment $AB$ would be a nice place to start.

Answer 1

You can try with the equation of circle $$x^2+y^2=1$$ and you can put $A$ insisde it, so $A(a,0)$ (we can assume $y=0$) where $a<1$ and $B(b,c)$ outside of it, so $b^2+c^2>1$. Suppose $b\ne a$. Now the equation of line $AB$ is $$y={c\over b-a} (x-a)$$ so the intersection points are solution to this equation $$\Big((b-a)^2+c^2\Big)x^2-2ac^2x+a^2c^2-(b-a)^2=0$$ Now all you have to check if the discriminant $D$ is greater than $0$:

\begin{align} {D\over 4} &= a^2c^4- \Big(a^2c^2-(b-a)^2\Big)\Big((b- a)^2+c^2\Big)\\ &= (b-a)^2 \Big((b-a)^2 +\underbrace{(1-a^2)}_{>0}c^2\Big)\\ &> 0 \end{align}

and we are done.